e to the x Implicit Derivative

[Jason76]

\(\displaystyle f(x) = e^{x/y} = 3x - y\)

\(\displaystyle f(x) = \ln(e^{x/y}) = \ln(3x - y)\)

\(\displaystyle f(x) = \dfrac{x}{y} = \ln(3x - y)\) - Quotient rule on left (once doing the derivative) and what on the right?

\(\displaystyle f'(x) = ?\)

 

[lookagain]

Drop the "f(x) = " parts to the left on the lines. You're doing implicit derivatives after all.

\(\displaystyle e^{x/y} = 3x - y\)

\(\displaystyle \ln(e^{x/y}) = \ln(3x - y)\)

\(\displaystyle \dfrac{x}{y} = \ln(3x - y)\) - Quotient rule on left (once doing the derivative) and what on the right?

\(\displaystyle ?\)

.

\(\displaystyle \dfrac{y(1) \ - \ x(y')}{y^2} \ = \ \dfrac{3 \ - \ y'}{3x \ - \ y} \ \ \ \ \ \ \ \ \ \ \ The \ \ right-hand \ \ side \ \ follows \ \ from \ \ the \ \ rule \ \ that \ \ d[ln(u)] \ = \ (du)/u. \)



Incomplete. This is so you can work further on it.